You have found the following ages (in years) of all 4 tigers at your local zoo: $ 7,\enspace 17,\enspace 2,\enspace 3$ What is the average age of the tigers at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 4 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{7 + 17 + 2 + 3}{{4}} = {7.3\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $-0.3$ years $0.09$ years $^2$ $17$ years $9.7$ years $94.09$ years $^2$ $2$ years $-5.3$ years $28.09$ years $^2$ $3$ years $-4.3$ years $18.49$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.09} + {94.09} + {28.09} + {18.49}} {{4}} $ $ {\sigma^2} = \dfrac{{140.76}}{{4}} = {35.19\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{35.19\text{ years}^2}} = {5.9\text{ years}} $ The average tiger at the zoo is 7.3 years old. There is a standard deviation of 5.9 years.